Convergence Test (Comparison) Questions

I need some help on doing convergence tests (comparisons I believe) on some Ʃ sums.

I have three, they are:

1. Ʃ [ln(n)/n^2] from n=1 to ∞.

I tried the integral test but was solved to be invalid (that is, cannot divide by infinity). Therefore I believe it to be a comparison test but don't know what to compare it to.

2. Ʃ [(3n+2)/(n^3+1)] from n=0 to ∞

I solved the whole problem starting at n=1 and stating that Ʃ1/n^2 and Ʃ1/n^3 were larger and convergent, therefore the sum is convergent. I realised the n=0 part later and now cannot solve it cause you cannot divide by 0; that is, 1/0^2 or 1/0^3. I tried an index shift yet it still is invalid.

3. Ʃ [(2n+1)/n^2] from n=1 to ∞

I don't know what to compare this to. I need a value smaller than 1/n^2 which diverges (as sum diverges) but 1/n^3 converges. I don't know what to do.

Staff: Mentor

I tried the integral test but was solved to be invalid (that is, cannot divide by infinity). Therefore I believe it to be a comparison test but don't know what to compare it to.

Why would you want to "divide by infinity"?
A comparison can be useful - you'll need something "between" 1/n (-> sum is not convergent) and 1/n^2 (->sum is convergent, but it is not larger than your expression)

2. Ʃ [(3n+2)/(n^3+1)] from n=0 to ∞

I solved the whole problem starting at n=1 and stating that Ʃ1/n^2 and Ʃ1/n^3 were larger and convergent, therefore the sum is convergent. I realised the n=0 part later and now cannot solve it cause you cannot divide by 0; that is, 1/0^2 or 1/0^3. I tried an index shift yet it still is invalid.

For convergence, finite numbers of initial summands are irrelevant. It is sufficient if your comparison works for n>2, n>1 billion or any other number.

3. Ʃ [(2n+1)/n^2] from n=1 to ∞

I don't know what to compare this to. I need a value smaller than 1/n^2 which diverges (as sum diverges) but 1/n^3 converges. I don't know what to do.

Help would be much appreciated.

The sum over 1/n^2 converges. You can split this into two series, both of them are easier to evaluate afterwards.